Chapter 14: Mathematical Reasoning Exercise 14.5

Let p: ‘If x is a real number, such that x³ + 4x = 0, then x is 0.’

q: x is a real number, such that x³ + 4x = 0

r: x is 0

(i) We assume that q is true to show that statement p is true and then show that r is true.

Therefore, let statement q be true.

Hence, x³ + 4x = 0

x (x² + 4) = 0

x = 0 or x² + 4 = 0

Since x is real, it is 0.

So, statement r is true.

Hence, the given statement is true.

(ii) By contradiction, to show statement p to be true, we assume that p is not true.

Let x be a real number, such that x³ + 4x = 0 and let x ≠ 0.

Hence, x³ + 4x = 0

x (x² + 4) = 0

x = 0 or x² + 4 = 0

x = 0 or x² = -4

However, x is real. Hence, x = 0, which is a contradiction since we have assumed that x ≠ 0.

Therefore, the given statement p is true.

(iii) By the contrapositive method, to prove statement p to be true, we assume that r is false and prove that q must be false.

∼r: x ≠ 0

Clearly, it can be seen that

(x² + 4) will always be positive.

x ≠ 0 implies that the product of any positive real number with x is not zero.

Now, consider the product of x with (x² + 4)

∴ x (x² + 4) ≠ 0

x³ + 4x ≠ 0

This shows that statement q is not true.

Hence, it proved that

∼r ⇒ ∼q

Hence, the given statement p is true.

The given statement can be written in the form of ‘if then’ is given below.

If a and b are real numbers, such that a² = b², then a = b.

Let p: a and b are real numbers, such that a² = b²

q: a = b

The given statement has to be proved false. To show this, two real numbers, a and b, with a² = b² , are required, such that a ≠ b.

Let us consider a = 1 and b = – 1

a² = (1)²

= 1 and

b² = (-1)²

= 1

Hence, a² = b²

However, a ≠ b

Therefore, it can be concluded that the given statement is false.

Let p: If x is an integer and x² is even, then x is also even.

Let q: x is an integer and x² is even.

r: x is even

By the contrapositive method, to prove that p is true, we assume that r is false and prove that q is also false.

Let x is not even.

To prove that q is false, it has to be proved that x is not an integer or x² is not even.

x does not even indicate that x² is also not even.

Hence, statement q is false.

Therefore, the given statement p is true.

(i) Let q: All the angles of a triangle are equal.

r: The triangle is an obtuse angled triangle.

The given statement p has to be proved false.

To show this, the required angles of a triangle should not be obtuse angles.

We know that the sum of all the angles of a triangle is 180⁰. Therefore, if all three angles are equal, then each angle measures 60⁰, which is not obtuse.

In an equilateral triangle, all angles are equal. However, the triangle is not an obtuse-angled triangle.

Hence, it can be concluded that the given statement p is false.

(ii) The given statement is

q: The equation x² – 1 = 0 does not have a root lying between 0 and 2.

This statement has to be proven false.

To show this, let us consider

x² – 1 = 0

x² = 1

x = ± 1

One root of equation x² – 1 = 0, i.e., the root x = 1, lies between 0 and 2.

Therefore, the given statement is false.

(i) The given statement p is false.

By the definition of the chord, it should intersect the circle at two distinct points.

(ii) The given statement q is false.

The centre will not bisect that chord which is not the diameter of the circle.

In other words, the centre of a circle only bisects the diameter, which is the chord of the circle.

(iii) The equation of an ellipse is,

If we put a = b = 1, then we get

x² + y² = 1, which is an equation of a circle.

Hence, the circle is a particular case of an ellipse.

Therefore, statement r is true.

(iv) x > y

By a rule of inequality,

-x < – y

Hence, the given statement s is true.

(v) 11 is a prime number.

We know that the square root of any prime number is an irrational number.

Therefore, √11 is an irrational number.

Hence, the given statement t is false.